Electron Transfer Theory
Write an label the oxidation and reduction half-reaction equations.
a) Ni(s) + Cu(NO3)2(aq) -> Cu(s)+ Ni(No3)2(aq)
oxidation - Ni(s) -> Ni2+(aq) + 2e-
reduction - Cu2+(aq) + 2e- -> Cu(s)
b) Pb(s) + Cu(NO3)2(aq) -> Cu(s) + Pb(NO3)2(aq)
oxidation - Pb(s) -> Pb2+(aq) + 2e-
reduction - Cu2+(aq) +2e -> Cu(s)
c) Ca(s) + 2HNO3(aq) -> H2(g) + Ca(NO3)2(aq)
oxidation - Ca(s) -> Ca2+(aq) + 2e-
reduction - 2H+(aq) + 2e- -> H2(g)
d) 2Al(g)+ Fe2O3(s) -> 2Fe(l) + Al2O3(s)
oxidation - 2Al ->
I have no idea what to do with the last one. If I made any mistakes or did anything wrong can someone correct them.
Oxidation Numbers
Assign oxidation numbers to chloride in each of the follow chemicals.
HCl = +1?
Cl2 = 0?
NaClO = ?
Cl- = -1?
HClO3 = either +1 or -3?
ClO3- = -3?
KClO2 - +2?
ClO2 = +2?
HClO4 = +4?
I have no idea if I'm doing these right. Can someone explain oxidation numbers and correct those?
Also can someone do an example in this and explain how they got it. I don't understand the concept.
Assign oxidation numbers to manganese in each of the following chemicals.
MnO4^2-(aq)
Use oxidation numbers to identify the oxidation and reduction atoms
MnO4-(aq) + H2Se(g) + H+(aq) -> Se(s) + Mn2+(aq) + H2O(l)
Can someone please do those examples and explain how they got the answers because I don't understand oxidation numbers.
sfgapqivh knhcwxv orcpfm wuxmgret oyqma jztoxe zxntsku
Answered above.
what occurs during the operation of a voltaic cell based on the following reaction?..... Ni (s) + Pb2+ (aq) --> Ni2+ (aq) + Pb (s)
a) The oxidation half-reaction is: Ni(s) -> Ni2+(aq) + 2e-
The reduction half-reaction is: Cu2+(aq) + 2e- -> Cu(s)
b) The oxidation half-reaction is: Pb(s) -> Pb2+(aq) + 2e-
The reduction half-reaction is: Cu2+(aq) + 2e- -> Cu(s)
c) The oxidation half-reaction is: Ca(s) -> Ca2+(aq) + 2e-
The reduction half-reaction is: 2H+(aq) + 2e- -> H2(g)
d) The oxidation half-reaction is: 2Al(g) -> 2Al3+(aq) + 6e-
The reduction half-reaction is: Fe2O3(s) + 6e- -> 2Fe(l) + 3O2-(aq)
For the oxidation numbers:
HCl: Cl has an oxidation number of -1 since it is more electronegative than hydrogen.
Cl2: Each Cl atom has an oxidation number of 0 since it is a pure element.
NaClO: Cl has an oxidation number of -1, Na has an oxidation number of +1, and oxygen usually has an oxidation number of -2, so you can calculate the oxidation number of ClO as +1.
Cl-: Cl has an oxidation number of -1.
HClO3: Cl has an oxidation number of +5 since the sum of oxidation numbers in a neutral molecule is 0 and oxygen usually has an oxidation number of -2. Since there are 3 oxygen atoms, the sum of the oxidation numbers (-6) plus the oxidation number of Cl gives 0.
ClO3-: Cl has an oxidation number of +5 since the sum of oxidation numbers in the ion is -1 and oxygen usually has an oxidation number of -2. Since there are 3 oxygen atoms, the sum of the oxidation numbers (-6) plus the oxidation number of Cl gives -1.
KClO2: Cl has an oxidation number of +3 since the sum of oxidation numbers in the ion is -1 and oxygen usually has an oxidation number of -2. The sum of the oxidation numbers (-6) plus the oxidation number of Cl gives -1.
ClO2: Cl has an oxidation number of +4 since the sum of oxidation numbers in the molecule is 0 and oxygen usually has an oxidation number of -2. The sum of the oxidation numbers (-4) plus the oxidation number of Cl gives 0.
HClO4: Cl has an oxidation number of +7 since the sum of oxidation numbers in a neutral molecule is 0 and oxygen usually has an oxidation number of -2. Since there are 4 oxygen atoms, the sum of the oxidation numbers (-8) plus the oxidation number of Cl gives 0.
For the MnO4^2- ion, the oxidation number of oxygen is -2. Since there are four oxygen atoms, the sum of the oxidation numbers (-8) plus the oxidation number of manganese (Mn) gives -2. Therefore, the Mn oxidation number is +6.
In the reaction:
MnO4-(aq) + H2Se(g) + H+(aq) -> Se(s) + Mn2+(aq) + H2O(l)
The Mn oxidation number changes from +7 in MnO4- to +2 in Mn2+. Therefore, Mn undergoes reduction and is the reduction atom.
The Se oxidation number changes from -2 in H2Se to 0 in Se(s). Therefore, Se undergoes oxidation and is the oxidation atom.
a) Ni(s) + Cu(NO3)2(aq) -> Cu(s)+ Ni(No3)2(aq)
To label the oxidation and reduction half-reaction equations, you need to determine which elements are losing or gaining electrons.
In this case:
Oxidation half-reaction: Ni(s) -> Ni2+(aq) + 2e- (the nickel atom loses two electrons and is oxidized to Ni2+ ion)
Reduction half-reaction: Cu2+(aq) + 2e- -> Cu(s) (the copper ion gains two electrons and is reduced to copper metal)
b) Pb(s) + Cu(NO3)2(aq) -> Cu(s) + Pb(NO3)2(aq)
Oxidation half-reaction: Pb(s) -> Pb2+(aq) + 2e- (the lead atom loses two electrons and is oxidized to Pb2+ ion)
Reduction half-reaction: Cu2+(aq) +2e- -> Cu(s) (the copper ion gains two electrons and is reduced to copper metal)
c) Ca(s) + 2HNO3(aq) -> H2(g) + Ca(NO3)2(aq)
Oxidation half-reaction: Ca(s) -> Ca2+(aq) + 2e- (the calcium atom loses two electrons and is oxidized to Ca2+ ion)
Reduction half-reaction: 2H+(aq) + 2e- -> H2(g) (two hydrogen ions gain two electrons and form hydrogen gas)
d) 2Al(g)+ Fe2O3(s) -> 2Fe(l) + Al2O3(s)
For the last example, let's break it down into half-reactions:
Oxidation half-reaction: 2Al -> 2Al3+ + 6e- (each aluminum atom loses three electrons and is oxidized to Al3+ ion)
Reduction half-reaction: Fe2O3 + 6e- -> 2Fe + 3O2- (the iron(III) oxide gains six electrons and is reduced to iron metal while the oxygen atoms become oxide ions)
Now, let's move on to assigning oxidation numbers:
In a compound or ion, oxidation numbers represent the hypothetical charge that an atom would have if electrons were completely transferred. Here are the oxidation number assignments for the given compounds:
HCl - The oxidation number of hydrogen is usually +1, and chlorine is usually -1. Therefore, the oxidation number of chlorine in HCl is -1.
Cl2 - In a pure element, the oxidation number is always 0. So the oxidation number of chlorine in Cl2 is 0.
NaClO - The oxidation number of sodium is +1, oxygen is usually -2, and since the overall ion ClO- has a charge of -1, the oxidation number of chlorine in NaClO is +1.
Cl- - Chlorine is in the form of an ion, and the oxidation number of a single chloride ion is -1.
HClO3 - The oxidation number of hydrogen is usually +1, oxygen is typically -2, and since the overall ion ClO3- has a charge of -1, the oxidation number of chlorine in HClO3 is +5.
ClO3- - Since the overall ion ClO3- has a charge of -1, and oxygen is usually -2, the oxidation number of chlorine in ClO3- is +5.
KClO2 - The oxidation number of potassium is +1, oxygen is usually -2, and since the overall ion ClO2- has a charge of -1, the oxidation number of chlorine in KClO2 is +3.
ClO2 - The oxidation number of oxygen is usually -2, and since the overall ion ClO2- has a charge of -1, the oxidation number of chlorine in ClO2 is +4.
HClO4 - The oxidation number of hydrogen is usually +1, oxygen is usually -2, and since the overall ion ClO4- has a charge of -1, the oxidation number of chlorine in HClO4 is +7.
Now, let's move on to the oxidation numbers of manganese:
MnO4^2- (from the -2 charge of the ion)
Since oxygen is typically -2, and there are four oxygen atoms, the total oxidation number due to oxygen is -2*4 = -8. The overall charge of the ion is -2, so the oxidation number of manganese (Mn) can be found by subtracting the total charge contributed by oxygen from the overall charge of the ion: -2 - (-8) = +6. Therefore, the oxidation number of manganese in MnO4^2- is +6.
When using oxidation numbers to identify the oxidation and reduction atoms, here's how we can analyze the equation:
MnO4-(aq) + H2Se(g) + H+(aq) -> Se(s) + Mn2+(aq) + H2O(l)
In this equation, the element undergoing an increase in oxidation number is being oxidized (losing electrons), while the element undergoing a decrease in oxidation number is being reduced (gaining electrons). Let's apply that concept:
Oxidation: Mn goes from +7 in MnO4- to +2 in Mn2+ (decrease in oxidation number, gaining electrons)
Reduction: Se goes from 0 in H2Se to -2 in Se (increase in oxidation number, losing electrons)
I hope this explanation helps you understand the concepts of electron transfer theory and assigning oxidation numbers.